Question: Solve for $x$, ignoring any extraneous solutions: $\dfrac{x^2 - 2x}{x + 10} = \dfrac{-20x - 80}{x + 10}$
Answer: Multiply both sides by $x + 10$ $ \dfrac{x^2 - 2x}{x + 10} (x + 10) = \dfrac{-20x - 80}{x + 10} (x + 10)$ $ x^2 - 2x = -20x - 80$ Subtract $-20x - 80$ from both sides: $ x^2 - 2x - (-20x - 80) = -20x - 80 - (-20x - 80)$ $ x^2 - 2x + 20x + 80 = 0$ $ x^2 + 18x + 80 = 0$ Factor the expression: $ (x + 8)(x + 10) = 0$ Therefore $x = -8$ or $x = -10$ However, the original expression is undefined when $x = -10$. Therefore, the only solution is $x = -8$.